Here’s a procedure for you to know how to descompose polynomials of any kind and how and when to use each of the different techniques.
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What is the factorial decomposition of a polynomial
Just like the decomposition of a number is to express it as a product of prime factors, such as for example:
To decompose a polynomial into factors is to express that polynomial as a product of factors, where this time the factors are either polynomials of degree 1 or irreducible polynomials of any degree.
The polynomials of degree 1, as we saw in the first lesson, are polynomials where the sum of the exponents of their variables is equal to 1 and can be simply a variable, if it is a polynomial with only one term or it can have two terms, where one of them will be of degree 0 (a number) and another of degree 1.
In general, they have this shape:
Where “a” is the coefficient that multiplies x and “b” is the number that forms the term zero degree. These are examples of grade 1 polynomials:
Irreducible polynomials are those that have no real roots and therefore cannot be broken down, as for example:
If we equal this polynomial to zero we solve the resulting equation, we will obtain two complex solutions, so this polynomial would be irreducible and therefore, it is also a factor when it comes to breaking down polynomials.
An example of decomposition of a polynomial into factors would be this polynomial:
Put it this way:
I mean, as a product of factors.
Let’s see how this is done in the next section.
How to break down a polynomial into factors
To break down a polynomial into factors, follow the procedure below and perform each of the following steps in this order, if possible:
- Extract common factor, if possible.
- If it is not possible to extract a common factor, it is necessary to identify if it is a remarkable product and if it is, write the polynomial as a remarkable product in its undeveloped form.
- If it is not possible to perform two previous steps, then decompose by Ruffini’s rule. This is the last resort you have to resort to when you want to break down an equation, as it is the least direct of all.
On the other hand, to break down a grade 2 polynomial, as long as it is not a remarkable product, the fastest way is to equal the polynomial to zero and get its roots.
Then, we will obtain binomials of the form (x-a) corresponding to each root and we will express the second degree polynomial as the product of the two binomials.
Resolved polynomial decomposition exercises
Let’s take a closer look at how to apply each of these steps to break down a polynomial into factors in the following solved exercises.
I will explain step by step how to decompose the following polynomials.
Exercise resolved 1
We start by checking whether it is possible to obtain a common factor: can we get a common factor?
Yeah, you can get a common factor of an x:
We’re still breaking down the grade two polynomials. As we have seen before, we can express it as a difference of squares:
And apply the sum per difference formula:
The polynomial would be factored in.
Exercise resolved 2
Is it possible to take out a common factor? Yes, we can get common factor to x²
Now we look at the parenthesis, in which we have a subtraction of two terms, can we convert it into one of these squares?
Yes, because x raised to 4 the is square of x squared and 1 squared equals 1:
We apply the sum per difference formula:
And now in the first parenthesis we can’t do anything else, since it’s an irreducible polynomial (you can check it by trying to get its roots), but in the third parenthesis we have a subtraction of squares again, so we apply the formula of addition by difference again:
Fully factored in.
Exercise solved 3
Can we get a common factor? Yes, we can draw a common factor of x:
In the parenthesis we have a second-degree polynomial that we will factorize by obtaining its roots. To do this we equal the polynomial to zero:
And we get their solutions, which correspond to the roots of the polynomial:
To each root corresponds a binomial of the form (x-a):
So we express the second degree polynomial as the product of these two binomials, leaving the initial factored polynomial:
Exercise solved 4
In this case, we can’t get a common factor or identify any notable product, so we factored by Ruffini.
The independent term is a 6, so we will have to try the following numbers: 1,-1, 2, -2, 3 and -3.
We’ve got some left:
And the factored polynomial is:
Exercise completed 5
Neither can we take out a common factor, nor apply any formula from a remarkable product, so we factor in the Ruffini rule.
The independent term is 1, so we’ll have to try 1 and -1. It’s left to us:
At this point, we can’t continue factoring until we have a grade 1 polynomial left (if you try to continue, you only have to try 1 and -1 and you won’t have a 0 left in the last row). Therefore, the remaining grade 2 polynomial is an irreducible polynomial and corresponds to a factor.
The factored polynomial remains:
Considerations to be taken into account
In a polynomial expressed as a multiplication of factors, its degree will be the sum of the degrees of each of the factors.
Therefore, make sure that the factorization of the polynomial has the same degree as the original polynomial.